MySQLi PHP结果数组不兼容？

Question

我正在一个项目上，代码中包含以下代码段：

<div id= "searchresults" style="min-height: 10vh; height: auto; background-color:white; display: block;">
<?php
//Makes connection
include 'dxbase.php';
$conn = new mysqli($servername, $username, $password, $dbname);

if(isset($_POST["search"])){
$tempvardf = '%'.$_POST['search'].'%'; #############
//Defines SQL command to bring up data about the products matching the set search
$sql = "SELECT id, name, price, location FROM products WHERE name LIKE ?;"; #'%$_POST[search]%'
//Create a prepared statement
$stmt = mysqli_stmt_init($conn);
//Prepare the prepared statement
if (!mysqli_stmt_prepare($stmt, $sql)) {
    echo "SQL STATEMENT FAILED... OOF";
} else {
    //Bind paramaters to placeholder
    mysqli_stmt_bind_param($stmt, "s", $tempvardf);
    //Run paramaters inside database
    mysqli_stmt_execute($stmt);
    $result = mysqli_stmt_get_result($stmt);

    while ($row = mysqli_fetch_assoc($result)) {
        echo $row['name']. "<br>";
        if(isset($row['path'])){
        echo "<img src=". $row['path'].">";
        } else {
            echo "can't access image path";
        }
    }
    echo "And that's it!";
}



} else {

}


?>
</div>

当我搜索时，出现了数据库中产品的名称，但是当我尝试回显图像时，它告诉我它无法访问图像路径。

图像路径在数据库中，如下所示证据：
，但是，我想知道哪里出了问题以及为什么它们没有显示...如此处所示：

尽管我尽了最大的努力和研究时间，但我仍然陷于困境...

Answer 1

您如何在Select语句中选择图像的路径？您的$ result中没有字段“路径”。